A thin rod lies along x from 0 to L = 1.2 m. Its linear density is λ(x) = λ₀(1 + ax/L), with λ₀ = 0.24 kg m⁻¹ and dimensionless a = 0.7. (a) Integrate to find its mass and centre-of-mass coordinate. (b) Where should a single vertical support be placed to balance the horizontal rod in a uniform gravitational field? Explain why L/2 is not correct.

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Mass 0.3888 kg; centre x_cm = 0.6519 m from x = 0. Place the support at x_cm, to the right of L/2. A short element dx has mass dm = λ(x)dx. Because density varies, replacing the rod by a uniform rod would incorrectly give equal masses to equal lengths. Integrate M = ∫₀ᴸ λ₀(1 + ax/L) dx = λ₀L(1 + a/2) = 0.3888 kg. The integral has units of mass because linear density is multiplied by length. The first moment is ∫₀ᴸ x dm = λ₀L²(1/2 + a/3). Divide this by M to obtain x_cm = L(1/2 + a/3)/(1 + a/2). Substitution gives x_cm = 0.6519 m. A single support at this position lies under the line of action of the total weight, so the gravitational torque about it vanishes. The centre is beyond L/2 because the right-hand side is denser, yet it remains within the rod. Setting a = 0 recovers L/2, a useful check on the nonuniform-density calculation.

Centre of mass of a nonuniform rod · set 2

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